Question: Simplify the following expression: $y = \dfrac{-2x^2+19x- 45}{-2x + 9}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-2)}{(-45)} &=& 90 \\ {a} + {b} &=& &=& {19} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $90$ and add them together. The factors that add up to ${19}$ will be your ${a}$ and ${b}$ When ${a}$ is ${9}$ and ${b}$ is ${10}$ $ \begin{eqnarray} {ab} &=& ({9})({10}) &=& 90 \\ {a} + {b} &=& {9} + {10} &=& 19 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-2}x^2 +{9}x) + ({10}x {-45}) $ Factor out the common factors: $ x(-2x + 9) - 5(-2x + 9)$ Now factor out $(-2x + 9)$ $ (-2x + 9)(x - 5)$ The original expression can therefore be written: $ \dfrac{(-2x + 9)(x - 5)}{-2x + 9}$ We are dividing by $-2x + 9$ , so $-2x + 9 \neq 0$ Therefore, $x \neq \frac{9}{2}$ This leaves us with $x - 5; x \neq \frac{9}{2}$.